Math Problem Statement

Determine whether the lines L₁: x = 3 + 2t, y = 4 - 2t, z = 1 - 3t and L₂: x = 1 + 4s, y = 3 - s, z = 4 + 5s are parallel, skew, or intersecting.

Solution

To determine whether the lines are parallel, skew, or intersecting, let's analyze the parametric equations of the lines:

Line L1L_1:

x=3+2t,y=42t,z=13tx = 3 + 2t,\quad y = 4 - 2t,\quad z = 1 - 3t where tt is the parameter.

Line L2L_2:

x=1+4s,y=3s,z=4+5sx = 1 + 4s,\quad y = 3 - s,\quad z = 4 + 5s where ss is the parameter.

Step 1: Check if the lines are parallel

For two lines to be parallel, their direction vectors must be scalar multiples of each other. The direction vector for L1L_1 is derived from the coefficients of tt: d1=2,2,3\vec{d_1} = \langle 2, -2, -3 \rangle The direction vector for L2L_2 is derived from the coefficients of ss: d2=4,1,5\vec{d_2} = \langle 4, -1, 5 \rangle Now, check if d1\vec{d_1} and d2\vec{d_2} are scalar multiples of each other. If they are not, the lines are not parallel.

Clearly, there is no constant kk such that d1=kd2\vec{d_1} = k \vec{d_2}, so the lines are not parallel.

Step 2: Check if the lines are intersecting

For the lines to intersect, there must be values of tt and ss that make the coordinates (x,y,z)(x, y, z) the same for both lines. This means we solve the system of equations: 3+2t=1+4s3 + 2t = 1 + 4s 42t=3s4 - 2t = 3 - s 13t=4+5s1 - 3t = 4 + 5s

Let's solve this system of equations to check if there are solutions for tt and ss.

Solving the system:

From the first equation: 3+2t=1+4s2t4s=2t=2s13 + 2t = 1 + 4s \quad \Rightarrow \quad 2t - 4s = -2 \quad \Rightarrow \quad t = 2s - 1

Substitute t=2s1t = 2s - 1 into the second equation: 42(2s1)=3s4 - 2(2s - 1) = 3 - s 44s+2=3s4 - 4s + 2 = 3 - s 64s=3s3s=3s=16 - 4s = 3 - s \quad \Rightarrow \quad 3s = 3 \quad \Rightarrow \quad s = 1

Substitute s=1s = 1 into t=2s1t = 2s - 1: t=2(1)1=1t = 2(1) - 1 = 1

Step 3: Verify with the third equation

Substitute s=1s = 1 and t=1t = 1 into the third equation: 13(1)=4+5(1)1 - 3(1) = 4 + 5(1) 13=4+5291 - 3 = 4 + 5 \quad \Rightarrow \quad -2 \neq 9 This is a contradiction, so the lines do not intersect.

Step 4: Conclusion

Since the lines are neither parallel nor intersecting, they must be skew.

Thus, the correct answer is: skew\boxed{\text{skew}}

Would you like further clarification on any part of the solution?

Here are 5 related questions to explore:

  1. What conditions must hold for two lines to be parallel in 3D space?
  2. Can two lines be skew in 2D space, or does this concept only apply to 3D?
  3. How can we geometrically visualize skew lines in 3D space?
  4. What methods are used to find the shortest distance between two skew lines?
  5. How do direction vectors influence the nature of the relationship between two lines?

Tip: When checking if lines are parallel, focus on their direction vectors. If these vectors are proportional, the lines are guaranteed to be parallel.

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Math Problem Analysis

Mathematical Concepts

3D Geometry
Lines in Space
Vector Analysis

Formulas

Direction vector for a line: (change in x, change in y, change in z)
Intersection conditions: Set x, y, z values of both lines equal and solve for parameters

Theorems

Condition for parallelism: Direction vectors must be scalar multiples
Condition for intersection: System of parametric equations must have a solution

Suitable Grade Level

Grade 11-12 or College Level